\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}\frac{1}{\frac{\sqrt{k} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}{1 \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1}{2}\right)}}}double f(double k, double n) {
double r161860 = 1.0;
double r161861 = k;
double r161862 = sqrt(r161861);
double r161863 = r161860 / r161862;
double r161864 = 2.0;
double r161865 = atan2(1.0, 0.0);
double r161866 = r161864 * r161865;
double r161867 = n;
double r161868 = r161866 * r161867;
double r161869 = r161860 - r161861;
double r161870 = r161869 / r161864;
double r161871 = pow(r161868, r161870);
double r161872 = r161863 * r161871;
return r161872;
}
double f(double k, double n) {
double r161873 = 1.0;
double r161874 = k;
double r161875 = sqrt(r161874);
double r161876 = 2.0;
double r161877 = atan2(1.0, 0.0);
double r161878 = r161876 * r161877;
double r161879 = n;
double r161880 = r161878 * r161879;
double r161881 = r161874 / r161876;
double r161882 = pow(r161880, r161881);
double r161883 = r161875 * r161882;
double r161884 = 1.0;
double r161885 = r161884 / r161876;
double r161886 = pow(r161880, r161885);
double r161887 = r161884 * r161886;
double r161888 = r161883 / r161887;
double r161889 = r161873 / r161888;
return r161889;
}



Bits error versus k



Bits error versus n
Results
Initial program 0.5
rmApplied div-sub0.5
Applied pow-sub0.4
Applied frac-times0.4
rmApplied clear-num0.4
Final simplification0.4
herbie shell --seed 2020047
(FPCore (k n)
:name "Migdal et al, Equation (51)"
:precision binary64
(* (/ 1 (sqrt k)) (pow (* (* 2 PI) n) (/ (- 1 k) 2))))